#### Prime divisor 7

There is another way of testing for divisibility by powers of primes
other than 2 and 5. We illustrate with the prime divisor 7.
This method uses the fact that 7 divides 2*10 + 1 = 21.
Start with the numeral for the number you want to test. Chop off the
last digit, double it, and subtract that from the rest of the number.
Continue this until you get a one-digit number. The result is 7, 0, or
-7, if and only if the original number is a multiple of 7.

**Example:**

123471023473

--> 12347102347 - 2*3 = 12347102341

--> 1234710234 - 2*1 = 1234710232

--> 123471023 - 2*2 = 123471019

--> 12347101 - 2*9 = 12347083

--> 1234708 - 2*3 = 1234702

--> 123470 - 2*2 = 123466

--> 12346 - 2*6 = 12334

--> 1233 - 2*4 = 1225

--> 122 - 2*5 = 112

--> 11 - 2*2 = 7.

Here's another explanation for this rule, contributed by **Dr. Mitteldorf**:

When you separate the last digit from the rest of a number, it's like
writing the number in the form 10x+y, where × is the number with the
last digit lopped off and y is the last digit.

When you double the last digit and subtract from the number formed by
the rest of the digits, you're computing x-2y. So the question becomes:
why is it that whenever 10x+y is a multiple of 7, x-2y is also, and vice versa?

If x-2y is divisible by 7, then certainly 10*(x-2y) is divisible by 7. That's 10x-20y. And 21y is always divisible by 7. So the sum, 10x-20y+21y=10x+y, is also divisible by 7.

A little thought reveals that it also works the other way: If 10x+y
is divisible by 7, and 21y is always divisible by 7, then 10x-20y must
be divisible by 7. But this can be written 10(x-2y), and certainly the 10 isn't divisible by 7, so the x-2y must be.

- Doctor Mitteldorf, The Math Forum

#### Prime divisor 13

An analogous method works for 13, using the fact that 13 divides
4*10 - 1 = 39. Start with the numeral for the number you
want to test. Chop off the last digit, double it, double it again, and
**add** that to the rest of the number. Continue this
until you get a number smaller than 40. The result is 13, 26, or 39 if
and only if the original number is a multiple of 13.
**Example:**

160518960524

--> 16051896052 + 4*4 = 16051896068

--> 1605189606 + 4*8 = 1605189638

--> 160518963 + 4*8 = 160518995

--> 16051899 + 4*5 = 16051919

--> 1605191 + 4*9 = 1605227

--> 160522 + 4*7 = 160550

--> 16055 + 4*0 = 16055

--> 1605 + 4*5 = 1625

--> 162 + 4*5 = 182

--> 18 + 4*2 = 26

#### Other prime powers up to 100 (multipliers at most 12)

Similar methods can be given for other prime powers, depending on
finding a small multiple of them which ends in 1 or 9. We given the
following table of prime powers up to 100 which have multipliers at
most 12. This covers all of them with the exception of only
73, 83, and 97:

#### Other bases

If a numeral is written in a base R other than 10, similar rules
can be devised for divisibility by prime powers. There are two main
cases: when the prime divides R, and when it does not. When it does,
looking at the last few digits of the numeral works. When it does not,
all the digits of the numeral will come into play.
The first kinds of tests will depend on the prime power factorizations
of R^k - 1 and R^k + 1 for positive integers k. The alternate kinds of
tests described above can also be used, and the key here is to find
small multiples of the divisor which end in 1, 01, 001, and so on, or
which end in one or more digits (R-1).